SVD : Why right singular matrix is written as transpose The SVD is always written as,
A = U Σ V_Transpose

The question is,
Why is the right singular matrix written as V_Transpose?
I mean lets say,
W = V_Transpose
and then write SVD as
A = U Σ W
SVD Image credit: https://youtu.be/P5mlg91as1c
Thank you
 A: It's written as a transpose for linear algebraic reasons.
Consider the trivial rank-one case $A = uv^T$, where $u$ and $v$ are, say, unit vectors. This expression tells you that, as a linear transformation, $A$ takes the vector $v$ to $u$, and the orthogonal complement of $v$ to zero. You can see how the transpose shows up naturally.
This is generalized by the SVD, which tells you that any linear transformation is a sum of such rank-one maps, and, what's more, you can arrange for the summands to be orthogonal.
Specifically, the decomposition
$$
A = U\Sigma V^T = \sum_{i = 1}^k \sigma_i u_i v_i^T
$$
says that, for any linear transformation $A$ on $\mathbb{R}^n$ for some $n$ (more  generally, any compact operator on separable Hilbert space), you can find orthonormal sets $\{v_i\}$ and $\{u_i\}$ such that

*

*$\{v_i\}$ spans $\ker(A)^{\perp}$.


*$A$ takes $v_i$ to $\sigma_i u_i$, for each $i$.
A special case of this is the spectral decomposition for a positive semidefinite matrix $A$, where $U = V$ and the $u_i$'s are the eigenvectors  of $A$---the summands $u_i u_i^T$ are rank-one orthogonal projections. For Hermitian $A$, $U$ is "almost equal" to $V$---if the corresponding eigenvalue is negative, one has to take $u_i = -v_i$ so that $\sigma_i \geq 0$.
A: $V^T$ is the Hermitian transpose (the complex conjugate transpose) of $V$.
$V$ itself holds the right-singular vectors of $A$ that are the (orthonormal) eigenvectors of $A^TA$; to that extent: $A^TA = VS^2V^T$. If we wrote $W = V^T$, then $W$ would no longer represent the eigenvectors of $A^TA$.
Additionally, defining the SVD as: $A = USV^T$ allows us to directly use $U$ and $V$ to diagonalise the matrix in the sense of $Av_i = s_iu_i$, for $i\leq r$ where $r$ is the rank of $A$ (i.e. $AV = US$). Finally using $USV^T$ also simplifies our calculation in the case of a symmetric matrix $A$ in which case $U$ and $V$ will coincide (up to a sign) and it will allows us to directly link the singular decomposition to eigen-decomposition $A = Q \Lambda Q^T$. Just to be clear: "yes, using $V^T$ instead of $W = V^T$ is a bit of convention" but is a helpful one.
A: My answer is much dumber than the others...

lets say, W = V_Transpose
and then write SVD as A = U Σ W

with that you are asking the reader to memorize one more variable ($W$) but for a simple expression as $V^T$ is just not worth it, IMO.
